Advertisements
Advertisements
प्रश्न
Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ?
उत्तर
Let the corresponding sides of similar triangles be s1 and s2.
Let A1 and A2 be their corresponding areas.
s1 : s2 = 4 : 7 ......[Given]
∴ `"s"_1/"s"_2= 4/7` ......(i)
by theorem of areas of similar triangles,
`"A"_1/"A"_2 = "s"_1^2/"s"_2^2`
`"A"_1/"A"_2 = ("s"_1/"s"_2)^2`
`"A"_1/"A"_2 = (4/7)^2` ......[From (i)]
`"A"_1/"A"_2 = 16/49`
∴ Ratio of areas of similar triangles = 16 : 49
APPEARS IN
संबंधित प्रश्न
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`
The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.
In the following figure RP: PK= 3:2, then find the value of A(ΔTRP):A(ΔTPK).
In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that :

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`
In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.
- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.
In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.
`"A(∆ ABD)"/"A(∆ ADC)"`
In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `("A(Δ ABC)")/("A(Δ DCB)")` = ?
In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.
`(A(∆ABD))/(A(∆ABC))`
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.
\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.
if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.
Areas of two similar triangles are in the ratio 144: 49. Find the ratio of their corresponding sides.
In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?
If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
A(ΔPBC) = `1/2 xx square xx square`
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`